Question

A parallelogram has sides with lengths of `18` and `4`. If the parallelogram's area is `48`, what is the length of its longest diagonal?

Answer 1

`sqrt(340+48sqrt(5)) ~~21.15`

Explanation:

The problem is to find `AC` in the image above, given
`AB=CD=18`
`AD=BC=4`
`"Area"_(ABCD)=48`

As the area of a parallelogram is given by the product of its base and height, we have `AE*CD=48`. Thus

`AE = 48/(CD)=48/18=8/3`

As `triangleAED` is a right triangle, we may apply the Pythagorean theorem to find that `AE^2+ED^2=AD^2`. Substituting in our known values for `AE` and `AD`, we get

`(8/3)^2+ED^2=4^2`

`=> ED=sqrt(16-64/9) = sqrt(80/9)=(4sqrt(5))/3`

As `triangleAEC` is also a right triangle, we can apply the Pythagorean theorem once more to get `AE^2+EC^2=AC^2`. As `EC = ED+CD`, we can once again substitute in our known values:

`AC^2 = AE^2+(ED+CD)^2`

`= (8/3)^2+((4sqrt(5))/3+18)^2`

`=340+48sqrt(5)`

Then, taking the square root, we get our final result:

`AC = sqrt(340+48sqrt(5)) ~~21.15`